Therefore, at point (2, 17), the slope of the tangent is 18.

[NoLimits]

Hello,

For this question I managed to find an answer but I am not sure if what the question means is to plug in the x-value and find the slope, or first find the derivative of the function, and THEN solve for the slope using f'(x)=10x+5h-2 (what I got for the derivative of the function). If someone could help explain the question I would appreciate it.

Homework Statement

For each function, use the definition of the derivative at a point to find the slope of the tangent at the indicated point.

Homework Equations

For point (2, 17): f(x) = 5x^2-2x+1

The Attempt at a Solution

f'(x) = lim_h→0 \frac{f(x+h)-f(x)}{h}
= lim_h→0 \frac{[5(2+h)^2-2(2+h)+1]-[5(2)^2-2(2)+1]}{h}
= lim_h→0 \frac{[5(4+4h+h^2)-4-2h+1]-[20-3]}{h}
= lim_h→0 \frac{20+20h+5h^2-3-2h-20-3}{h}
= lim_h→0 (18+5h)
= 18 + 5(0)
= 18

 

[Mark44]

Hello,

For this question I managed to find an answer but I am not sure if what the question means is to plug in the x-value and find the slope, or first find the derivative of the function, and THEN solve for the slope using f'(x)=10x+5h-2 (what I got for the derivative of the function). If someone could help explain the question I would appreciate it.

Homework Statement

For each function, use the definition of the derivative at a point to find the slope of the tangent at the indicated point.

Homework Equations

For point (2, 17): f(x) = 5x^2-2x+1

So f(2) = 5(2)² - 2(2) + 1 = 20 - 4 + 1 = 17

The Attempt at a Solution

f'(x) = lim_h→0 \frac{f(x+h)-f(x)}{h}
= lim_h→0 \frac{[5(2+h)^2-2(2+h)+1]-[5(2)^2-2(2)+1]}{h}

In the line above you aren't finding f'(x), which is implied by the line above it. What you are doing is finding f'(2). Your first equation should have been
$$f'(2) = lim_h→0 \frac{f(2+h)-f(2)}{h} $$

= lim_h→0 \frac{[5(4+4h+h^2)-4-2h+1]-[20-3]}{h}
= lim_h→0 \frac{20+20h+5h^2-3-2h-20-3}{h}
= lim_h→0 (18+5h)
= 18 + 5(0)
= 18

Aside from what I already said, this looks fine. What your answer means is that at the point (2, 17), the slope of the tangent line is 18.